Q. The amount of work done in blowing a soap bubble such that its diameter increases from $d$ to $D$ is ($S =$ surface tension of solution)
A) $\pi\left( D^{2} - d^{2} \right)S$
B) $2\pi\left( D^{2} - d^{2} \right)S$
C) $4\pi\left( D^{2} - d^{2} \right)S$
D) $8\pi\left( D^{2} - d^{2} \right)S$
Explanation
(b) Change in surface area
$2 \times 4\pi\lbrack(D/2)^{2} - (d/2)^{2}\rbrack = 2\pi(D^{2} - d^{2})$
Work done = surface tension $\times$ change in area
$= S \times 2\pi(D^{2} - d^{2})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.